\begin{proof}
Given a preference game over player set $[n]$, with the preference lists specified as a set of values $Q_{ij}$ for all $i, j \in [n]$: $Q_{ij} = $ the number of players $k$ such that $j \geqprefer_i k \geqprefer_i i$.

Define a game as follows, in which we will find a personalized equilibrium. 
\begin{itemize}
\item The set of players = $\{p_1, \ldots, p_n\}$
\item $S_i$ (the set of strategies for player $p_i$) = $\{s_{ij}: Q_{ij} > 0\}$
\item $E_i$ = the set  of edges for player $p_i$ = $\{ \{s_{ij}, s_{jj}\} \forall s_{ij} \in S_i, j \neq i\} \cup \{s_{ii}\}$
\item $u_i(\{s_{ij}, s_{jj}\})$ (the payoff to player $i$ for this edge) = $Q_{ij}$
\item $u_i(\{s_{ii}\}) = Q_{ii} \geq 1$
\end{itemize}

Notice that the degree of the game is preserved, and the number of edges defined is at most $n$ times the degree.

\subsubsection*{A personalized equilibrium maps to an equilibrium in the preference game.}
The map will be as follows: Suppose we are given weights $x_{ij}$ for each player $i$ and edge $\{s_{ij}, s_{jj}\}$, and $x_{ii}$ for player $i$ and edge $\{s_{ii}\}$. These weights form a personalized equilibrium.  We will set weights $w_{ij} = x_{ij}$ in the preference game.

To show this is an equilibrium in the preference game, we must show the following.
\begin{itemize}
\item For all $i$, $\sum_j w_{ij} = 1$. 

$\sum_j w_{ij} = \sum_j x_{ij} = 1$, since this is a valid solution to the personalized game.
\item For all $i,j$, $w_{ij} \leq w_{jj}$.

$w_{ij} = x_{ij} \leq x_{jj}$ (by the projection constraint for personalized equilibria), $x_{jj} = w_{jj}$.

\item $w$ is a lexicographically maximal weight assignment.

Suppose this is not true. Then, there exists another weight assignment $w'$ that is lexicographically larger than $w$. Let $w'$ be the lexicographically maximal such assignment. Thus, there exist $i, j$ such that $\sum_{k: Q_{ik} \geq Q_{ij}} w'_{ij} > \sum_{k: Q_{ik} \geq Q_{ij}} w_{ij}$. For this $i$, take a $j$ with the largest $Q_{ij}$ that meets this condition. By our definition of $j$ and the fact that $w'$ is lexicographically maximal, we know that for all $j'$ with $Q_{ij'} > Q_{ij}$, $\sum_{k: Q_{ik} \geq Q_{ij'}} w'_{ij} = \sum_{k: Q_{ik} \geq Q_{ij'}} w_{ij}$. Let $\delta = \sum_{k: Q_{ik} \geq Q_{ij}} w'_{ij} - \sum_{k: Q_{ik} \geq Q_{ij}} w_{ij} > 0$. Now consider the payoff to player $i$ in the personalized game.
\allowdisplaybreaks {
\begin{eqnarray*}
\textrm{Payoff to $i$} & = & \sum_l x_{il} Q_{il} \\
& = & \sum_{l: Q_{il} > Q_{ij}} x_{il} Q_{il} + \sum_{l: Q_{il} = Q_{ij}} x_{il} Q_{il} + \sum_{l: Q_{il} < Q_{ij}} x_{il} Q_{il} \\
& = & \sum_{l: Q_{il} > Q_{ij}} w_{il} Q_{il} +  Q_{ij} \sum_{l: Q_{il} = Q_{ij}} w_{il} + \sum_{l: Q_{il} < Q_{ij}} w_{il} Q_{il} \\
& = & \sum_{l: Q_{il} > Q_{ij}} w'_{il} Q_{il} +  Q_{ij} \sum_{l: Q_{il} = Q_{ij}} w_{il} + \sum_{l: Q_{il} < Q_{ij}} w_{il} Q_{il} \\
& = & \sum_{l: Q_{il} > Q_{ij}} w'_{il} Q_{il} +  Q_{ij} \left( (\sum_{l: Q_{il} = Q_{ij}} w'_{il}) - \delta \right) + \sum_{l: Q_{il} < Q_{ij}} w_{il} Q_{il} \\
& \leq & \sum_{l: Q_{il} > Q_{ij}} w'_{il} Q_{il} +  Q_{ij} \sum_{l: Q_{il} = Q_{ij}} w'_{il} - \delta Q_{ij} + \sum_{l: Q_{il} < Q_{ij}} w'_{il} Q_{il} \\
&& + \delta (Q_{ij} - 1) \\
& < & \sum_{l: Q_{il} > Q_{ij}} w'_{il} Q_{il} +  Q_{ij} \sum_{l: Q_{il} = Q_{ij}} w'_{il} - \delta (Q_{ij} - 1) \\
&& + \sum_{l: Q_{il} < Q_{ij}} w'_{il} Q_{il} + \delta (Q_{ij} - 1) \\
& = & \sum_{l: Q_{il} > Q_{ij}} w'_{il} Q_{il} +  Q_{ij} \sum_{l: Q_{il} = Q_{ij}} w'_{il} + \sum_{l: Q_{il} < Q_{ij}} w'_{il} Q_{il} \\
& = & \sum_l w'_{il} Q_{il} \\
\end{eqnarray*}
}
So player $i$ would do strictly better by playing $x = w'$, leading to
a contradiction.
\end{itemize}

\subsubsection*{An equilibrium in the preference game maps to a personalized equilibrium.}
Suppose we are given weights $w_{ij}$ forming an equilibrium in the preference game. We will set weights in the personalized game as follows. $x_{ie} = w_{ij}$ for player $i$ and edge $e=\{s_{ij}, s_{jj}\}$. $x_{ie} = w_{ii}$ for player $i$ and edge $e=\{s_{ii}\}$. 

To show this is a personalized equilibrium, we must show the following.

\begin{itemize}
\item For all $i$, $\sum_e x_{ie} = 1$. 

$\sum_e x_{ie} = \sum_j w_{ij} = 1$, since this is a valid weight assignment in the preference game.
\item For all $i, s_{jk} \in S_j$, $\sum_{e: s_{jk} \in e} x_{ie} \leq \sum_{e: s_{jk} \in e} x_{je}$.

If $j \neq k$, $\sum_{e: s_{jk} \in e} x_{ie} = 0 \leq \sum_{e: s_{jk} \in e} x_{ij}$. If $j=k$, $\sum_{e: s_{jj} \in e} x_{ie} = x_{ie'}$ where $e' = \{s_{ij}, s_{jj}\}$ $= w_{ij} \leq w_{jj} = x_{je''}$ where $e'' = \{s_{jj}\}$ $ = \sum_{e:s_{jj} \in e} x_{je}$.
\item $x$ is a best response in the personalized game for all players $i$.

Consider any other weight function $x'$ for the personalized game. Since there is a one-to-one mapping from defined edges to $i,j$ pairs in the preference game (including $i=j$), we can define a new weight function $w'$ in the preference game using the same rules as defined in the first half of this proof ($w'_{ij} = x'_{ij})$. We know that $w$ is lexicographically maximal for the preference game. Using the same reasoning as above, we get:

\begin{eqnarray*}
\textrm{Payoff to $i$ playing $x'$} & = & \sum_l x'_{il} Q_{il} \\
& = & \sum_{l: Q_{il} > Q_{ij}} x'_{il} Q_{il} + \sum_{l: Q_{il} = Q_{ij}} x'_{il} Q_{il} + \sum_{l: Q_{il} < Q_{ij}} x'_{il} Q_{il} \\
& = & \sum_{l: Q_{il} > Q_{ij}} w'_{il} Q_{il} +  Q_{ij} \sum_{l: Q_{il} = Q_{ij}} w'_{il} + \sum_{l: Q_{il} < Q_{ij}} w'_{il} Q_{il} \\
& < & \sum_{l: Q_{il} > Q_{ij}} w_{il} Q_{il} +  Q_{ij} \sum_{l: Q_{il} = Q_{ij}} w_{il} + \sum_{l: Q_{il} < Q_{ij}} w_{il} Q_{il} \\
& = & \sum_l w_{il} Q_{il} \\
& = & \textrm{Payoff to $i$ playing $x$} \\
\end{eqnarray*}
\end{itemize}

\end{proof}